
#1
Mar510, 08:13 PM

P: 16

when X is even number,it's easy to prove
but how about the condition which X is odd number? I have no idea of this 



#3
Mar510, 09:57 PM

P: 16





#4
Mar510, 11:25 PM

PF Gold
P: 1,930

how to prove √X is irrational number
So in other words...
[tex]\sqrt{x}[/tex] is irrational iff x=/=n^2 for n belonging to the integer set. 



#5
Mar510, 11:54 PM

P: 16





#6
Mar610, 12:49 AM

P: 234

Fundamental theorem of arithmetic. Assume p^2/q^2=x with gcd(p,q)=1, and see what has to divide what.




#7
Mar610, 06:33 PM

P: 16





#8
Mar610, 06:40 PM

P: 234

Greatest common divisor. If gcd(p,q)=1, it means the fraction p/q is in lowest terms.
Look at the proof for sqrt(2), and adapt it. Remember that "even" just means "is divisible by 2", so that if you're checking a number other than 2, you won't be thinking about "even" anymore. 



#9
Mar610, 06:50 PM

P: 1,784

Although right now you're probably more interested in just getting the right answer, you might want to check out the following Wikipedia entries:
http://en.wikipedia.org/wiki/Square_root http://en.wikipedia.org/wiki/Square_root_of_2 And for an interesting history of the discovery of irrational numbers look at http://en.wikipedia.org/wiki/Irrational_number 


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